MiScion Pty Ltd

18/08/2024

The wind classes for future reference. For design purposes our primary interest is the site reference pressure qzu, which we otherwise use with a surface pressure coefficient (Cpn) to get a surface pressure p=Cpn.qzu.

As they are ultimate strength values, they are the peak loads just prior to breaking or the actual breaking load depending on material. Deformation is permanent, it does not elastically recover after experience of ultimate strength loads. Ultimate strength essentially indicates end-of-life for the structure, it is no longer serviceable and needs to be repaired or replaced.

By contrast a purpose designed storm shelter would be serviceable during and after the severe storm event. Deformations would be elastic and would recover after the load is removed.

So with respect to wind class N1. At wind speeds of 26m/s the house remains serviceable and suitable to occupy. At the old permissible stress wind of 28m/s the deformation may still be elastic. At the ultimate strength wind speed of 34m/s [122km/h] the deformations are becoming permanent and the building may not be suitable to occupy during the storm event.

It should be noted that site wind speeds are derived from regional wind speeds typically recorded at airports/airfields (Terrain category 2), at a height of 10m above the ground. The regional wind speed for most of South Australia is 45 m/s [162 km/h].

So annual storm damage at wind speeds of 90km/h to 110 km/h not really acceptable, if these are wind speeds recorded at Adelaide airport or other weather observatory. If they are actual site/locational wind speeds then slightly more acceptable.

It is also to be noted that Bureau of meteorology issues severe weather warnings when wind average wind speed is 63km/h and when instantaneous wind speed is 90km/h. At around 63km/h trees sway and twigs break off, at around 90km/h trees will be uprooted. So it is a time to take shelter in a building that is serviceable at such wind speeds and clear from trees.

18/08/2024

The calculations in the previous posts were done using a combination of ATCalc, freemat, MS Excel and Jupyter(python) notebooks.

Large round off errors occurred using ATCalc as the calculations weren't continuous and linked. The rounded values of each stepwise calculation were used in each subsequent calculation, accumulating larger and larger differences.

Whilst the Calculations done using MS Excel, freemat and Jupyter using linked calculations, using the computers calculated result in each subsequent calculation, reducing the round error between calculations.

I use ATCalc as it is simple for random calculations. Freemat, and variants such as scilab, have an untidy interface after a few calculations. Whilst Jupyter using python, keeps the calculation sequence tidy and visible, the problem is python is cumbersome, need to import libraries, and may need more variables than other languages (or maybe that was another language). Whilst MS Excel, doesn't require variables, and calculations can be relatively free format.

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18/08/2024

As indicated in previous post I'm not sure why the version of AS1684 I have is is based on a roof load width (RLW) aligned with the roof slope rather the plan projected area, with respect to designing wall framing. Its cumbersome.

If I work out the weight of insulated timber roof structure, including the ceiling, I get a weight of 0.42kPa on plan area. Split in two then have ceiling about 0.27kPa on horizontal plane, whilst roof the raftered part is about 0.13kPa along slope. So allowing for 35 roof slope and transforming to horizontal get 0.16kPa for the roof plane. So in combination get 0.43kPa on horizontal plane, whilst for flat roof would be 0.4kPa. This is different than the whole 0.4kPa roof construction rotated 35 degrees which would give 0.49kPa on the horizontal. However, a roof with raked ceiling doesn't require ceiling joists and rafters and so would expect to be lighter.

The roof live load (LL=0.25kPa) is definitely defined on the plan projected area of the roof. Whilst wind load normal to face of the roof doesn't require magnitude transforming into vertical and horizontal components: the trig cancels and the magnitudes remain the same. It therefore doesn't make sense to define a RLW for the wall along the roof slope.

AS1684.1, which has been replaced by AS1720.?, uses 40 kg/m^2 for sheet metal roof mass and multiples by 0.01 to get kPa. That is 10/1000 where g=10 m/s^2 rather than g=9.81m/s^2, so conservative for gravity loading conditions but not wind uplift. Though in some places for the wall it uses 0.4kPa directly. It then use RLW*S to get supported area, where RLW is measured along roof slope (as defined in AS1684.2), and S is the appropriate spacing of the members. So why is RLW defined in such manner when not one of the loads relevant to the wall is defined so?

We cannot combine area loads or line loads if they are not similar in nature. So if deadload is along slope of roof, and live load on the horizontal, cannot calculate 1.2DL+1.5LL until convert to same nature. And not going to transform the LL because that is in the preferred form.

If consider LL only and its definition. Then if the maximum load width LW=7.5m when flat, then the same horizontal load width applies when at 35 degrees, and the roof length along slope can increase to 9.156m. However, the roof pitch and roof length doesn't matter, no calculations are required. The horizontal length is typically the distance between walls, or wall and ridge, the distance is a known starting point.

If consider the DL, then it is along the slope of the roof, so whilst it is 7.5m on the horizontal when a flat roof, it has to be 7.5m along the slope when rotated 35 degrees, the horizontal decreasing to 6.929m. With the load changing from 0.4kPa along horizontal for flat roof and 0.43kPa along horizontal for roof at 35 degrees, assuming that not rotating the ceiling. If rotate the ceiling then changes from 0.4kPa to 0.49kPa. So trig is required to calculate the loads. If RLW is along the slope, then can ignore the trig for the loads, but the user of the tables has to use trig to get the RLW.

But if using RLW along slope then the LL needs transforming from along the horizontal to along the slope, that is 0.25*cos(35)=0.20kPa. It reduces because the total load is going to be based on a longer length than the horizontal.

The structural model in AS1684.1 does not transform either load, there is no trig involved. For simplicity they appear to be taking the loads on the flat, and calculating the maximum load width, then declaring when roof rotated the load width is measured along the slope. The total load on the roof from both components of load does not change.

But strict application of the load types, the LL does change if simply rotating the roof and projected horizontal is decreasing. So on the flat we have 1.2DL+1.5LL=1.2*0.4+1.5*0.25=0.855kPa and load width 7.5m, so w=0.85*7.5=6.41kN/m along the wall.

If rotate 35 degrees. Then horizontal is 6.144, and LL contribution to wall is 1.5*0.25*6.144=2.30kN/m. Whilst the DL contribution is still 1.2*0.4*7.5=3.6kN/m, for a total of 2.3+3.6=5.90kN/m along the wall.

Alternatively transforming the DL, we get 1.2*0.49+1.5*0.25=0.96kPa, and w=0.96*6.144 = 5.90kN/m along the wall. OR transforming the LL, 1.2*0.4+1.5*0.2=0.78kPa, w=7.5*0.78=5.85 kN/m or without stepwise rounding errors w=5.90kN/m.

So can write load on wall as w=x*(1.2DL/cos(alpha) + 1.5*LL) or w=L*(1.2DL + 1.5LL*cos(alpha)) where is 'x' is the horizontal dimension or 'L' is length along the slope, and alpha is the roof elevation angle.

However if want to take advantage of the AS1684 span tables for the wall framing, then no trig' was used, and the load width is along the slope. So there is no change in load at the wall with the roof rotation. So total p=0.855kPa, and w=6.41kN/m at wall, as calculated above. But want to work with horizontal distances as its more convenient, and total at wall has to remain the same, that is W=po.L=p1.x and po/p1=x/L=cos(alpha), so p1=po/cos(alpha) for the whole load, not part of it. So p1=0.855/cos(35)=1.04kPa, and w=1.04*6.144=6.39kN/m or 6.41kN/m with reduced rounding errors. The load width now has to be less than 6.144m horizontal rather than 7.5m along rather.

So now instead of just using the roof mass to proportion the canopy width can use a more complete load case. So from previous example, house width 8.4m and load width to the wall is 4.2m, which is less than 6.144m. The load is 1.04*4.2=4.37kN/m and limit is 6.41kN/m so have reserve of 6.41-4.37=2.04kN/m.

The load from canopy is 1.2*0.2+1.5*0.25=0.61kPa, s=w/p=2.04/0.61=3.344m, and canopy width becomes 2*3.344 = 6.688m.

If we reduced house roof pitch to more typical of 22.5 degrees. Then 0.855/cos(22.5)=0.93kPa and x=6.929m. So load from house is 0.93*4.2 = 3.906kN/m, and reserve 6.41-3.906=2.50kN/m, and so s=2.50/0.61=4.10m, and canopy width 2*4.1=8.2m.

There are other load cases to consider like wind loading, but initially we want a quick check on maximum width to start with, and then reduce width as required. The span tables limit maximum tension in wall stud to 8.5kN, so we don't want wind load on wall stud to exceed this, but again few installed connections would have such capacity, so acceptable wind uplift load on wall studs much less than this.

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17/08/2024

The 7.5m roof load width (RLW) is defined measured parallel to the roof surface. This is generally inconvenient. Most handbooks of frame formula define uniformly distributed loads (UDL), distributed either horizontally or vertically, and this is because it is also easier to derive formula this way, it is also beneficial to look at the influence of a load in the horizontal and vertical directions. Our codes also define live loads distributed along the horizontal, 0.25kPa roof live load for example is distributed along the projected horizontal dimension not the actual length.

If we have stick of wood with a mass of say 5.94kg/m and it is 6m long, then its total mass is 35.64kg, and this acts through the centre of mass which is 3m from each end. The timber framing code limits roof pitch to 35 degrees, so if we rotate this stick of wood, through 35 degrees, it will still have a mass of 35.64kg acting vertically.

Now if we have rafters at 1.2m centres then the 0.25kPa will produce 0.30kN/m on each rafter. But this is not distributed along the length of the rafter, its distributed along the projected horizontal. It therefore cannot be combined with the deadload of the roof or self weight of the stick of wood, as they are based on different lengths.

The length of the horizontal is x = r.Cosθ = 6*cos(35)=4.915m and the vertical rise is y = 6*Sin(35) =3.441m. So the total live load (LL) is 4.915*0.3=1.47 kN not 6*0.3=1.8kN. So if we want to use the horizontal dimensions the self weight needs to be increased, so that when multiplied by 4.915 it gives the correct total weight. So we could calculate the total weight and distribute along the desired length. So have original length L, and new length x=L.Cosθ , so W=w.L=w1.x, therefore w/w1=x/L=Cosθ, and so w1=w/Cosθ, so the 5.94kg/m becomes 5.94/cos(35)=7.25kg/m distributed along the horizontal. Check: 7.25*4.915 = 35.63kg, which if remove some of the rounding error is acceptable.

Similarly don't want to work with 40 kg/sq.m or 0.4kPa distributed along length of rafter, want to work with horizontal distances between walls, not dimensions along roof slopes. So allowing for 35 degree roof pitch the load becomes 0.4/cos(35)=0.49kPa distributed along the horizontal, or 49 kg/sq.m. But now also working with horizonal distances so the 7.5m RLW, now becomes a load width of 6.144m on the horizontal.

Our flat verandah roof probably only a weight of 0.2kPa, with pitch less than 10 degrees. So converts to 0.2/cos(10)=0.20kPa so no real change.

So crudely the maximum load permitted along the wall is 6.144*0.49 = 3.01kN/m, which should match 7.5*0.4=3.00kN/m.

Typically the load width to a wall is half the distance between parallel walls, or half the width of the house. It could be more or less depending on how the structure is configured. But if symmetrical gable roof spanning between external walls, its half the width of the building at that point. So our building can have a maximum width of 2*6.144=12.29m, so if that is what we are starting with then we have no reserve for attaching a verandah.

So assume house has 2 rooms 3.6m wide and a 1.2m hallway between, so width is 2*3.6+1.2=8.4m, so load width to wall is 8.4/2=4.2m. Maximum permitted is 6.144m, so we have some reserve: 6.144-4.2=1.94m, which is certainly not the 6m or more people want to sling of their house roof and walls. But the verandah roof is only about 0.1 to 0.2 kPa roof weight. Where as our limit is based on 0.4 kPa (or 0.49kPa on horizontal).

So 4.2*0.49=2.06kN/m, and limit is 3kN/m, so 3-2.06=0.94kN/m. So canopy weight is 0.2kPa, and so s=w/p=0.94/0.2=4.7m. Check: 4.7*0.2=0.94kN/m

Now given that this is the load width of the canopy on the house wall, the canopy width can be twice this so 4.7*2 = 9.40m.

However, this is only considering one of many load cases concerned with the design of the house/veranda combination. Also need to consider live loading requirements, and wind loading. So would consider load combinations such as 1.2DL+1.5LL, and 0.9DL+WL. We also making an assessment on a defined RLW, and we don't know which load case is critical to determining such load width. Just making an assumption that the load is proportional to the weight, that everything else is constant and varies in similar manner.

All that basically done is determine that the weight of the verandah is not likely a critical issue if it uses lighter weight construction to the house. However the wind uplift will be critical, as will the resistance of the installed house connections.

So for wind class N1, qzu=0.69kPa, adopt Cpn=1.1, so pn=Cpn.qzu= 1.1*0.69=0.76kPa, and vertical, uplift component of wind for the RLW=7.5m, is LW=6.144m, so w=0.76*6.144=4.67kN/m, which is greater than the maximum 3kN/m of roof weight permitted, therefore there will be uplift at the roof and tie-down required.

But house likely designed for the 4.2m load width only not the maximum, so for w=0.76*4.2=3.19kN/m , and roof weight was only 2.06kN/m, so net uplift of 3.19-2.06=1.13kN/m, and from previous post the wall weight is less than this, so the weight of the concrete footing is important.

Longitudinal wind pressures on canopies vary between about 0.3 and 0.4, so even if transverse gets down to 0.2 such is not relevant. For narrow canopies the pressure coefficient can be around 1 to 1.5, which is slightly less than the eaves overhang at 1.6. But if we have a steep pitch and block the volume under then the coefficient can get still higher. A blocked volume is something like a large caravan parked under a carport: it is not a side wall or fence.

So house tie-down designed for 1.13kN/m, so assuming a nail in the footing is worth 1kN, and placed at every wall stud at 600mm centres, then it provides 1.67kN/m tie-down, but at 1.2m centres, it only provides 0.83kN/m resistance. Check: 1.13*1.2=1.36kN required at every wall stud carrying a rafter. An anchor bolt can likely provide at least 5kN at each wall stud, so 5/1.2=4.17kN/m. We thus have reserve capacity at the bottom plate. But not necessarily have adequate resistance between wall stud and bottom plate to provide reserve to attach a verandah.

So reserve capacity in house roof and wall tie-down connections needs to be checked before deciding on size of attached verandah. The bottom plate to slab connection, and wall stud to bottom plate connections are not readily accessible and therefore impractical to strengthen. It is therefore preferable to strengthen these connections when designing the house in the first instance.

The timber framing code limits tension in a wall stud to 8.5kN, but there a few timber framing connections which develop this capacity in the wall stud. The nominal, traditional connections, have less than 1kN capacity, and close enough to zero to be considered zero and none existent. Some codes require minimum connection capacity to be at least 50% of member capacity. So minimum connection would be 8.5/2=4.25kN, which would require use of framing brackets or steel strap to all connections not just nailing. For JD4, steel strap with 2 nails would give 3.5kN for wall stud to plate connection, whilst 3 nails would push up to 4.7kN capacity. whilst two nails through the plate into stud is worth only 0.17kN, and relatively useless.

Thus changing roof cladding from concrete roof tiles at 54.6kg/m² to steel roof cladding at 5kg/m² is going to make a big difference to the tie-down requirements in the house roof and wall framing. Its is not just a visual or decorative change. Similarly slinging an extra few metres of verandah roof from a house roof/wall structure is going to increase uplift on the house structure. The house connections likely don't have adequate reserve capacity, and therefore the verandah should be freestanding, unless want to consider the verandah a proper house renovation and strengthen the house wall structure. Which requires removing the internal lining from the wall frames, or removing bricks and creating openings from the outside to access the bottom plates.

Main issue here however is that we can convert our loads from working along the true length of the roof slope, to working on the projected horizontal length. Our live loads and wind uplift loads are on the horizontal, but our dead loads and self weights are along the true length of the element. Roof wind load also has a horizontal component, distributed on projected vertical height, which can be considered separately. Working with these horizontal distances can make some calculations easier. Our primary concern is maximum width of canopy can put on house wall.

Though should be careful with the loads. Whilst the standardised loads of 0.1 to 0.2kPa are along the slope of the roof, the 0.4kPa of the roof on the wall is already likely in a horizontal plane. So I'm not certain why AS1684 uses RLW aligned with slope to size wall framing.

Will take a closer look at that. But note ceiling is in horizontal plane, and roof structure sloping. So one part needs converting and the other doesn't.

17/08/2024

One of my dislikes of the timber framing code AS1684.2 is the structural model in AS1684.1 (now AS1720.?) has a formula which comprises of the term 2/W, which for a 2 story dwelling it doubles to 4/W. Where 'W' is the width of the building in metres. The intent of the formula is to convert the wall weight into a pressure distributed over the plan area of the house for some simple uplift calculations, that is convert to kPa=kN/m^2. Personally I consider it is a bad idea, and the load path should be properly followed from the roof to the wall, and the weight of the wall considered in terms of kN/m along the length of the wall.

Besides disliking converting to an area, the other issue I have is that the coefficient '2' has to have the units of kN/m. Given that AS1684 is for light timber framed construction, that is an heavy wall. Whilst most of our houses are brick veneer, the weight of the bricks are not relevant to AS1684. Bricks are not light weight cladding, and they do not contribute directly to the hold-down of the timber framing, even if they are connected by wall ties. The only contribution to hold-down provided by the masonry is the weight onto the footings. Given that we can only use 0.9 of the weight against uplift, the weight of a 2.4m high wall (110mm thick) is around 5.88 kN/m, thus greater than the coefficient of '2'.

Considering a timber framed wall from 90x45 pine, then the timber section weighs 2.23kg/m. So if we consider, 2.4m wall, a single stud, top plate, bottom plate and a noggin, and a 600mm segment, then the framing mass is 9.26kg, or 15.43 kg/m length of wall. If we line one side with 10mm plasterboard, the mass is 22.94kg or 38.23kg/m.

Then to turn to weight we multiply by acceleration due to gravity. For quick calculations that is 10m/s^2, or more refined 9.81 m/s^2, If concerned with live loads and dead loads, then the larger value is conservative, but if concerned with wind uplift, that would be unconservative. And no the 0.9G doesn't allow for such approximation, it allows for variation in density and dimensions, and consequently variation in section areas and volumes.

So at one extreme, 0.9G should allow for the 5th percentile weight of an element, whilst 1.2G should allow for 95th percentile weight, when we calculate using nominal values.

So we multiply by 9.81 to convert to Newtons [N] and divide by 1000 to convert to kilonewtons. This means the weight of a timber framed wall in brick veneer construction is 0.37kN/m and a long way from the coefficient of '2'.

If we clad the external side of the wall with corrugated steel, then wall weight is 0.48 kN/m. Or if use heavy CFC cladding, then may push it up to 1.05 kN/m. Given that typically have two walls want to distribute over the floor area, then we can consider doubling these values and getting close to the coefficient of '2'. But more than likely will be using a lighter version of CFC, bring wall weight down to 0.6kN/m.

So AS1684 is potentially over estimating the available hold-down resistance. First by adopting a high standardised weight for the wall, and secondly by distributing the load over the floor area.

I consider it far more preferable to distribute the wind load on the roof into reactions/actions at the walls in kN/m along the wall, and trace the load path down each wall. Noting that the available hold-down weight will vary with the height of the wall. So if the wall for brick veneer is reduced from 2400mm to 600mm high, then its weight is 0.14kN/m, and at 300mm high 0.10kN/m. We don't get the full weight until get to the bottom of the full height wall.

The segment of wall above a door or window, lacks weight for hold-down, so lintel has to transfer uplift into posts down into the footings. But we do have some small amount of weight there to provide some resistance to uplift.

The timber framing code indicates that a 90x45 MGP10 wall stud has a maximum roof load width of 7.5m. For simplicity assume a flat roof. Given Cpn=-1.1 and qzu=0.69kPa, then pn=1.1*0.69=0.76kPa, and w=0.76*7.5=5.69kN/m. Which is far greater than our 0.37kN/m for our wall framing.

Allowing for 0.4kPa roof mass, then that is 3kN/m along wall, of which can only consider 2.7kN/m as resistance, so net uplift is 5.69-2.7=2.99kN/m. And still don't have adequate uplift resistance in our wall framing, the weight of the footings is now important.

Considering attached verandahs again. The maximum load width for the wall is 7.5m, so combined load width from existing house and the attached verandah has to be less than 7.5m. Also preferably the house should be bigger than the verandah if the house footings and wall framing are to have reserve capacity for hold-down.

There is some potential for adjustments, as verandah roofs are lighter, and wind pressure coefficients (Cpn) are likely lower, though they may be the same.

The first check would be to use AS1684.2 as a simple check. If verandah not viable then need more detailed design to see if viable. If it appears viable using AS1684, then need more detailed check to ensure is viable. Using AS1684 it is not possible to check full viability of attached verandah, (unless more recent versions have more detailed information) . But it is possible to assess that really pushing the boundaries as to whether feasible or not.

Attached verandahs are house extensions and the existing house needs giving more consideration, before constructing such verandahs. Verandah manufacturers product brochures are not adequate for assessing suitability of the house.

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06/07/2024

From another perspective, not interested in wind speeds. Instead we have a strength grading system, grade 'A' reference pressure is qzu=0.69 kPa, grade 'AA' qzu=0.96kPa and grade 'AAA' qzu=1.5kPa.

We determine the design reference pressure for our project is qzu=0.95 kPa, therefore we have to select strength grade 'AA' for our components.

But if we want to put a 2 storey building on the site, then qzu=1.05kPa , and we have to select strength grade 'AAA' components for our project.

If we want to provide a general assessment for the site up to a 2 storey building then we declare the site is rated 'AAA'. But we can already see a lack of economy, as the 2 storey project only requires design for qzu=1.05kPa and we are selecting components rated for qzu=1.5kPa.

So if most of our projects on a site are less than 2 storeys, have a height smaller than the main single storey dwelling, then we don't want to be using components based on a 2 storey rating, we want a strength grade for our components which better matches the project.

So not modifying the classification of the site, rather simply finding an equivalent strength grade to allow selection of suitable components for a specific project.

So if happy using qzu=1.5kPa (AAA) to select components for all projects then can do that. Or can use the project specific design pressure of qzu=0.95kPa and design all components from first principles, or find an equivalent strength grade in this case 'AA' with qzu=0.96kPa, and select components based on this project specific strength grade. In this case grade 'AA' is close to our design requirements and is economical.

But qzu=0.95kPa and qzu=1.05kPa are not close to grade "AAA" for which qzu=1.5kPa, and therefore our simplified approach to design and component selection is not a benefit.

The simplified wind loading code AS4055 does not allow determination of project specific site/building reference pressure qzu, it only allows determination of wind class. Then from the wind class the upper limit on reference wind speed and pressure, and these may be higher than out project really needs. It is always based on a reference height of 7.25m.

Using AS1170.2, the reference wind speed Vzu, and design pressure qzu can be calculated and a wind class then assigned. These values are calculated for the actual height of the project, not a default height of 7.25m. But this is assignment of wind class is project specific. The assignment for the site has to be based on the reference height of 7.25m, allowing for a 2 storey building.

The whole point of AS4055 is to avoid using AS1170.2 and allow simplified selection of components, for buildings up to 2 storey's. The benefit of using AS1170.2 however is only for those situations where M[z,cat] varies for heights less than 10m, and that is for terrain category 2 and terrain category 1.

For terrain category 3 and 4, the benefit of using AS1170.2 is to know the value of qzu, and then know how close it is to the value for the assigned wind class, and decide whether to use the wind class or specific value of qzu.

If we use a specific bracket for our connections, all we really want to know is when we have to stop using it, and what we have to use in its place?

It should be noted that qzu is a reference pressure, and has to be converted into an component pressure using a pressure coefficient Cpn. So that design pressure is pn=Cpn.qzu. However, for a specific application we expect that Cpn is a constant, and therefore we only need to know qzu.

Basically if a wind speed map, or site assessment to AS4055 assigns wind class N3 or higher to your site, then there is likely benefit in getting an assessment to AS1170.2 and getting a value for qzu.

For wind class N1 and N2, whilst there maybe benefit in knowing qzu, it likely doesn't provide much benefit as the available components are likely better than needed for wind class N2, and smaller or otherwise more economical component not available. There maybe benefit if specifying two or more components, where a smaller qzu may allow one component. But if only using one, there isn't anything smaller, and cannot get rid of the component. So the decision to get an assessment to AS1170.2 will be once start selecting components, and it becomes apparent there maybe a possibility of using fewer. Where as for higher wind classes, seek AS1170.2 assessment, and get value of qzu, straight way.

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06/07/2024

Comparing AS4055 and AS1170.2 more directly. I have a site, using AS1170.2 the site reference wind spaces are Vzu[0]=37.88 m/s and Vzu[90]=39.87 m/s, assuming I know the orientation of the building and can determine a different wind speed parallel to the ridge line and transverse to the ridge line. But reducing this to a single wind speed then we get Vzu=39.87m/s for the site. This is determined for the average building height 3.822m, and the associated wind pressure is qzu=0.95kPa.

For wind class N1 Vzu=34 m/s and that for N2 Vzu=40m/s. So the site has to be classified as wind class N2. But council likely says it is wind class N3.

If I change the average building height to 7.25m, then Vzu=41.15m/s and qzu=1.02 kPa. For wind class N3 Vzu=50m/s and qzu=1.5kPa. Given that 41 exceeds 40, the site has to be classified as wind class N3. But 1.5/1.02 = 1.47 increase in pressure, and if consider height reduction then 1.5/0.95 = 1.58 increase in pressure over using AS1170.2 directly.

Now maybe consider that cannot use wind class N2 because didn't determine it at the correct height of 7.25m. But that is not so, the definition of the wind class is based purely on the wind speed.

The catch is the use of the simplified assessment method in AS4055, and the difference between site and product. When the site is classified we expect the classification is suitable for a 2 storey building.

But AS1170.2 indicates we can design the single storey building for wind class N2. If we then choose to do that, we cannot extend the building into a 2 storey building at some future date as that requires the building be N3. On the other hand all the lower storey bracing and tie-downs would need redesigning in any case. So we just reclassification the site/building combination to class N3 and select suitable members: which will have to change in any case for a 2 storey building.

Another situation is an attached carport or garden fence. Using the simplified method, get a wind classification suitable for a 2 storey dwelling. The fence is less than a storey in height, using AS1170.2 and adjusting for height instead of getting wind class N3, we get say wind class N2. Span tables and other design tables based on wind class N2 (qzu=0.96kPa) are suitable for our needs, as they are based on the pressure we expect to experience. So that fence which appeared unsuitable may well be suitable.

Similarly, if we adopt N2 for a verandah as that is less than 7.25m high, and assume house tie-downs assessed for wind class N2 in first instance, and we have reserve capacity in the existing connections. Then verandah is viable.

But if house designed for wind class N3 and all its tie-downs, then we have a double advantage. All the original connections installed to resist wind class N3 loads, we now assess tie-down requirements for the house to wind class N2, and now have greater reserve capacity available. Then we assess the extra loads imposed by the verandah to wind class N2. We therefore have potential for increased reserve available in the connections.

However a site classified as wind class N1 using AS4055 is more likely to have a higher wind speed when checked in detail to AS1170.2, this is because people using the latter likely to make a better assessment of available shielding in all 8 compass directions, and a consequential lack of shielding in at least one direction will result in wind speed being higher than Vzu=34m/s. At which point better to use AS1170.2 and ignore wind classification.

For ancillary structures, at wind classes N2 and above there is benefit to use AS1170.2 to allow for height. So potential to attach a wind class N1 verandah to a wind class N2 house.

Which likely explains why there isn't more damage to houses, when the tie-downs to the house are being ignored when installing attached verandahs. Again the extra strap around the rafter is little value, if the wall stud to top plate connection, the wall stud to bottom plate connection, and the bottom plate to slab connections are all inadequate to resist the extra wind load imposed by the attached verandah.

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